A slab of copper of thickness b = 8.000×10-1 mm is thrust into a parallel-plate

Question

A slab of copper of thickness b = 8.000×10-1 mm is thrust into a parallel-plate capacitor of C = 1.00×10-10 F of gap d = 6.0 mm, as shown in the figure; it is centered exactly halfway between the plates. If a charge q = 6.00×10-6 C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted?

How much work is done on the slab as it is inserted?

Is the slab pulled in or must it be pushed in? (You have only ONE try for this answer!)

The bold question is the one I cannot find the answer to.

Copper (c26p74) A slab of copper of thickness b 8.000x10 1 mm is thrust into a parallel-plate capacitor of C 1.00x10 10 Fof gap d 6.0 mm, as shown in the figure it is centered exactly halfway between the plates. If a charge q 6.00x10 6 C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? 1.15 You are correct. Previous Tries Your receipt no. is 152-4059 How much work is done on the slab as it is inserted? 0.0235 J Work could be positive or negative Submit Answer Incorrect. ries 4/10 Previous Tries pulled s the slab pulled in or must it be pushed in? You have only ONE try for this answer!) You are correct. Previous Tries Your receipt no. is 152-4216

Explanation / Answer

b = 8*10^-1 mm , C = 1*10^-10 F,

d = 6mm, q = 6*10^-6 C

C = Aeo/d

1*10^-10 = A*8.85*10^-12/0.006

A = 0.068 m^2

C1 = C2 = eo*A/(0.5(d-b))

1/Ctot = 1/C1 +1/C2

Ctot = eo*A/d-b

Ctot = 8.85*10^-12*0.068/(0.006 -0.0008)

Ctot= 1.157*10^-10 F

W = DU = Uf - Ui

= Q^2/2Ctot - Q^2/2Ci

= 0.5*(6*10^-6)^2((1/1.157*10^-10) - (1/10^-10))

W = -0.18 J

The system lost energy

The slab was sucked in

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